package leetcode

import kotlinetc.println

//https://leetcode.com/problems/matchsticks-to-square/

/**
Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:
Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.
Example 2:
Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks.
Note:
The length sum of the given matchsticks is in the range of 0 to 10^9.
The length of the given matchstick sort.getArray will not exceed 15.
 */
fun main(args: Array<String>) {

    makesquare(intArrayOf(5, 5, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3,2)).println()
}


//由于需要在所有的可能中找出一种符合要求的，可以考虑深度遍历
fun makesquare(nums: IntArray): Boolean {

    if (nums.size < 4) return false

    val C = nums.sum()

    if (C % 4 != 0) return false

    val sideLen = C / 4

    val visited = Array(nums.size) { false }

    val squares = Array(4) { 0 }

    nums.sortDescending()
    return makesquareRecursive(nums, squares, sideLen, 0)

}

var count = 0
fun makesquareRecursive(nums: IntArray, squares: Array<Int>, sideLen: Int, pos: Int): Boolean {

    val size = nums.size
    if (pos == size && squares.all { it == sideLen }) return true


    var result: Boolean
    val side = nums[pos]
    for (j in 0 until 4) {
        if (squares[j] + side <= sideLen) {
            squares[j] += side

            result = makesquareRecursive(nums, squares, sideLen, pos + 1)
            if (result) return result

            squares[j] -= side
        }
    }

    return false
}

